Question 243826
log(4x-15) = log(x) + log(x+4)


becomes:


log(4x-15) = log(x*(x+4)) because log(a*b) = log(a) + log(b)


In this case, a was equal to x and b was equal to (x+4)


if log(4x-15) = log(x*(x+4)) then:


4x-15 = x*(x+4) which becomes:


4x-15 = x^2 + 4x


subtract 4x from both sides of this equation to get:


x^2 + 4x - 4x = -15 which becomes:


x^2 = -15


take the square root of both sides of this equation to get:


x = +/- sqrt(-15)


since you are taking the square root of a negative number, and the result of that is not real, then the number of real solutions that this equation has is 0.